Finding restriction of the ultraproduct that behaves like $\mathbb{Z}$

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Let $\mathcal{A}= \prod_{n \in \mathbb{N}} \mathcal{A}_n /\mathcal{U}$, where $\mathcal{A}_n=(\{0, 1, \dots, n\},<)$ and $\mathcal{U}$ is a non-principal ultrafilter of $\mathbb{N}$.

Can we find some $S \subseteq A$ such that $(S,<^\mathcal{A}|_S)$ is isomorphic to $\mathbb{Z}$?


I see that the smallest element of $\mathcal{A}$ is $[0,0, \dots]$, and the greatest is $[0,1,2, \dots]$.

My idea was to take the following approach:

  1. Find an explicit scheme for constructing the $A$-successor of an element $\neq [0,0,\dots]$

  2. Find an explicit scheme for constructing the $A$-predeccessor of an element $\neq [0,1,2,\dots]$

  3. Find an explicit element of $A$ such that, no matter how many times we apply 1 or 2, we never hit $[0,0, \dots]$ or $[0,1,2,\dots]$

  4. Let $S$ be the closure of the element from 3 under the operations from 1 and 2, and prove $S$ has the desired properties.

I know that non-prinicpal means the intersection of all sets in $\mathcal{U}$ is just the empty set. I know this is equivalent to saying that there is no $n$ such that $n$ belongs to every $U \in \mathcal{U}$.

I'm still new to ultrafilters so I would like to do this from scratch as much as possible (i.e. without relying on any theorems that I haven't learned yet. I know Los Theorem but that's about it.)